Let $A$ be an abelian free group that is finitely generated, and let $B\subset A$ be a subgroup of $A$ such that $A/B$ is a torsion group. Show that $rank(A)=rank(B)$.
From the hypothesis, I know that $A=<x_1,..., x_n>$, so $rank(A)=n$. Since the set $\{x_1, ..., x_n\}$ clearly generates $B$, then I have that $rank(B)\leq n$. Now, the idea that I have is to show that if $S\subset B$ generates $B$, then it also generates $A$, but I don't know if this is true, and I don't know how to prove it. I was trying to do this, and here's what I got:
Since $A/B$ is a torsion group, for $a\in A$, there exists a natural number $m$ such that $ma\in B$, so $ma$ can be written as a linear combination of elements in $S$. (I don't know if this is useful).
Any hint would be very appreciated! Thank you!
$\langle X\rangle$
for $\langle X\rangle$ and use$\operatorname{rank}(C)$
for $\operatorname{rank}(C)$. $\endgroup$ – Shaun Dec 11 '18 at 4:04